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Friday, March 25, 2016

Irrational number

Prove that the number $\newcommand{\lcm}{{\rm lcm}}$

$$\mathcal{I}=\sum_{n=1}^{\infty} \frac{1}{\lcm (1, 2, \dots, n)}$$

is irrational.

Solution


Assume it's rational, say $\frac{p}{q}$. Define $p_k$ and $q_k$ such that $\frac{p_k}{q_k} = \sum \limits_{n=1}^k \dfrac{1}{\lcm(1, 2, \ldots, n)}$. We then have the lower bound

$$\frac{p}{q} - \frac{p_k}{q_k} \geq \frac{1}{qq_k} \geq \frac{1}{q \lcm(1, 2, \ldots k)}$$

We also have

$$\frac{p}{q} - \frac{p_k}{q_k} = \sum_{n=k+1}^{\infty} \frac{1}{\lcm(1, 2, \ldots, n)} = \left(\frac{1}{\lcm(1, 2, \ldots k)}\right)\left(\sum_{n=k+1}^{\infty} \frac{\lcm(1, 2, \ldots k)}{\lcm(1, 2, \ldots n)}\right)$$

where the last sum can be made arbitrarily small by setting $k+1$ to be a large prime. In particular, it can be made smaller than $\frac{1}{q}$, which induces a contradiction with our lower bound.

The exercise can also be found in AoPS.com

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